Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $z = \dfrac{4r - 24}{5r + 20} \times \dfrac{r^2 - 16}{-2r^2 - 12r + 80} $
Solution: First factor out any common factors. $z = \dfrac{4(r - 6)}{5(r + 4)} \times \dfrac{r^2 - 16}{-2(r^2 + 6r - 40)} $ Then factor the quadratic expressions. $z = \dfrac {4(r - 6)} {5(r + 4)} \times \dfrac {(r - 4)(r + 4)} {-2(r - 4)(r + 10)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {4(r - 6) \times (r - 4)(r + 4) } {5(r + 4) \times -2(r - 4)(r + 10) } $ $z = \dfrac {4(r - 4)(r + 4)(r - 6)} {-10(r - 4)(r + 10)(r + 4)} $ Notice that $(r - 4)$ and $(r + 4)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {4\cancel{(r - 4)}(r + 4)(r - 6)} {-10\cancel{(r - 4)}(r + 10)(r + 4)} $ We are dividing by $r - 4$ , so $r - 4 \neq 0$ Therefore, $r \neq 4$ $z = \dfrac {4\cancel{(r - 4)}\cancel{(r + 4)}(r - 6)} {-10\cancel{(r - 4)}(r + 10)\cancel{(r + 4)}} $ We are dividing by $r + 4$ , so $r + 4 \neq 0$ Therefore, $r \neq -4$ $z = \dfrac {4(r - 6)} {-10(r + 10)} $ $ z = \dfrac{-2(r - 6)}{5(r + 10)}; r \neq 4; r \neq -4 $